Armorikam

2020-10-26

The integrals converge. Evaluate the integrals without using tables

$\int}_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}$

2abehn

Skilled2020-10-27Added 88 answers

Step 1: Given that

The integrals converge. Evaluate the integrals without using tables

$\int}_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}$

Step 2: Solve

We have,

${\int}_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}}dr={\int}_{0}^{1}\frac{4r}{\sqrt{1-{\left({r}^{2}\right)}^{2}}}dr$

Substitute,

${r}^{2}=t$

2r.dr=dt

$dr=\frac{dt}{2r}$

Plugging all the values into the integral we obtain,

$\int}_{0}^{1}\frac{4r}{\sqrt{1-{\left({r}^{2}\right)}^{2}}}dr{\int}_{0}^{1}\frac{4r}{\sqrt{1-{t}^{2}}}\times \frac{dt}{2r$

$=2{\int}_{0}^{1}\frac{dt}{\sqrt{1-{t}^{2}}}$

$=2{\left[{\mathrm{sin}}^{-1}\left(t\right)\right]}_{0}^{1}$

$=2({\mathrm{sin}}^{-1}\left(1\right)-0)$

$=2{\mathrm{sin}}^{-1}\mathrm{sin}\left(\frac{\pi}{2}\right)$

$=2\times \frac{\pi}{2}$

$=\pi$

The integrals converge. Evaluate the integrals without using tables

Step 2: Solve

We have,

Substitute,

2r.dr=dt

Plugging all the values into the integral we obtain,

Jeffrey Jordon

Expert2021-11-08Added 2607 answers

Answer is given below (on video)

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